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Four squares of sides $x$ cm each are cut off from the four corners of a square metal sheet having side $100$ cm. The residual sheet is then folded into an open box which is then filled with a liquid costing Rs. $x^2$ with $cm^3$. The value of $x$ for which the cost of filling the box completely with the liquid is maximized, is

  1. $100$
  2. $50$
  3. $30$
  4. $10$
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Answer will be 30 Lets Try to see how it came

First of all it is a metal sheet of square shape of length 100 cm now 4 squares of side x cm  are removed from four corners so now length of remaining metal sheet will be 100-2x and when it is folded it will become cuboid in which length will  be 100-2x and breadth 100-2x and height will be x now Volume of this cuboid is= L X B X H=(100-2x)(100-2x)x=(100-2x)^2x cm^3 now cost of 1 cm^3 is x^2 so for entire cost will be (100-2x)^2x^3 now in order to find maximum perform differentiation. UV rule of differentitaion ….. it will be -4x+3(100-2x)=0 so -4x+300-6x=0  so 10x=300 so  x is 30.

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